Answer to Question #217338 in Algebra for Jay

Question

Assume that the sum of 100 integers n₁, n₂, .... is greater than 100. So

A. All the numbers n₁, n₂, .... have to be greater than or equal to 1.

B. If one of the numbers n₁, n₂, .... is lower or equal to -100, so at least one of the remaining numbers must be greater than 2

C. If between the numbers Solutions of the inequality [(4x²-3x-1)^1/2]>= 2x-3 are?

Answer 1

A. False. Counterexample: let $n_1=-2<1, n_2=5, n_3=n_4=...=n_{100}=1$

$$S_{100}=-2+5+1+1+...+1=3+98(1)=101>100$$

B. True. Let $n_1=-100, n_2\leq 2, n_3\leq2,..., n_{100}\leq 2$

$$S_{100}\leq-100+99(2)=98<100$$

We have contradiction.

Therefore If one of the numbers n₁, n₂, .... is lower or equal to -100, so at least one of the remaining numbers must be greater than 2.

C.

$$(4x^2-3x-1)^{1/2}\geq2x-3$$

$$4x^2-3x-1\geq0=>(x-2)(4x+1)\geq0$$

$$x\leq-0.25\ or\ x\geq2$$

If $x\leq-0.25,$ then $2x-3<0.$

If $x\geq2,$ then $2x-3>0$

$$4x^2-3x-1\geq(2x-3)^2$$

$$4x^2-3x-1\geq4x^2-12x+9$$

$$9x\geq10$$

$$x\geq \dfrac{10}{9}$$

Solution of the inequality

$$x\in(-\infin, -0.25]\cup [2,\infin )$$